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Mumz [18]
3 years ago
8

a radio station can broadcasts in all directions to a distance of 60 mi. what is the area in which the station can be heard

Mathematics
2 answers:
zaharov [31]3 years ago
4 0

Answer:

Area = PI * radius ^ 2

Area = PI * 60^2

Area = PI * 3,600

Area = 11,309.73 square miles

Step-by-step explanation:

frosja888 [35]3 years ago
4 0
<h2>Answer:</h2>

<u>The station can be heard to </u><u>11304 miles</u>

<h2>Step-by-step explanation:</h2>

Area of circle is given by

<h3>A = π * r² </h3>

Putting the values

A = 3.14 * 60²

Aera = 11304 miles

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Step-by-step explanation:

First convert 2.5*10^-5 into decimal form

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2.5*10^-5 = 0.000025

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3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and
lisov135 [29]

Answer:

(a) 0.50928

(b) 0.857685.

Step-by-step explanation:

We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb i.e.;                                                         \mu = 160 lb  and \sigma = 27.5 lb

(A) We know that Z = \frac{X - \mu}{\sigma} ~ N(0,1)

Let X = randomly selected pilot  

If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)

P(150 < X < 201) = P(X < 201) - P(X <= 150)

P(X < 201) = P( \frac{X - \mu}{\sigma} < \frac{201 - 155}{29.2} ) = P(Z < 1.57) = 0.94179

P(X <= 150) = P( \frac{X - \mu}{\sigma}  < \frac{150 - 155}{29.2} ) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749

                                                                                                   = 0.43251

Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .

(B) We know that for sampling mean distribution;

           Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);

 P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)

P(X bar < 201) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{201 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < 9.84) = 1 - P(Z >= 9.84)

                                                                                  = 0.999995

P(X bar <= 150) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{150 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < -1.07) = 1 - P(Z < 1.07)

                                                                                   = 1 - 0.85769 = 0.14231

Therefore,  P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.

C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.

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